On 12/13/2011 01:58 AM, Jesse Phillips wrote:
On Sun, 11 Dec 2011 19:40:42 +0100
Mafi<m...@example.org>  wrote:

void f(ref const(int)* a, const(int)* b) {a = b; }
void main() {
    immutable(int)* a;
    auto b = (new int[](5)).ptr;
    f(a, b);
    //if this compiles I have just assigned a mutable pointer
    //to an immutable one
}

With the above definition using inout, such a call would be
disallowed.

Just gave this code a try:

void f(ref const(int)* a, const(int)* b) {a = b; }
void main() {
    const(int)* a;
    immutable(int)* b;
    auto c = (new int[](5)).ptr;
    f(a, c);
    f(b, c); //test.d(7): Error: cast(const(int)*)b is not an lvalue
}

Interesting. That is another bug that masks the one you were trying to exploit. :o) Consider:

void f(const ref int* a) { }
void main() {
    immutable(int)* a;
    f(a); // same error, but should compile
}

Anyway this does the same bad stuff as your example was trying to do and currently compiles:

void f(const(int)** a, const(int)* b) {*a = b; }
void main() {
    const(int)* a;
    immutable(int)* b;
    auto c = (new int[](5)).ptr;
    f(&a, c);
    f(&b, c); // now goes through
}






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